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March 8th, 2017, 11:02 AM   #1
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Solve trigonometry problem

Hello,
Please help. I have been given the question:

Find all solutions (in degrees) in the range −360° < θ < 360° such that cos(4θ) = cos(θ)

I have no clue how to even approach this question, can someone please help me?

Thanks in advance.

Last edited by skipjack; March 8th, 2017 at 11:22 AM.
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March 8th, 2017, 11:32 AM   #2
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The equation implies that 4θ = ±θ + k*360°, where k is an integer.
Hence θ is an integer multiple of 72° or 120°.
List the 13 solutions that lie in the specified interval.
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March 8th, 2017, 02:56 PM   #3
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Quote:
Originally Posted by Heath2017 View Post
Hello,
Please help. I have been given the question:

Find all solutions (in degrees) in the range −360° < θ < 360° such that cos(4θ) = cos(θ)

I have no clue how to even approach this question, can someone please help me?

Thanks in advance.
$\displaystyle \begin{align*} \cos{ \left( 4\,x \right) } &= \cos{ \left( x \right) } \\ 2\, \left[ \cos{ \left( 2\,x \right) } \right] ^2 - 1 &= \cos{ \left( x \right) } \\ 2\,\left[ 2\cos^2{ \left( x \right) } - 1 \right] ^2 - 1 &= \cos{ (x)} \\ 2\,\left[ 4\cos^4{(x)} - 4\cos^2{(x)} + 1 \right] - 1 &= \cos{(x)} \\ 8\cos^4{(x)} - 8\cos^2{(x)} + 2 - 1 &= \cos{(x)} \\ 8\cos^4{(x)} - 8\cos^2{(x)} - \cos{(x)} + 1 &= 0 \\ 8\cos^2{(x)}\,\left[ \cos^2{(x)} - 1 \right] - \cos{(x)} + 1 &= 0 \\ 8\cos^2{(x)}\,\left[ \cos{(x)} - 1 \right] \left[ \cos{(x)} + 1 \right] -1\,\left[ \cos{(x)} - 1 \right] &= 0 \\ \left[ \cos{(x)} - 1 \right] \,\left\{ 8\cos^2{(x)}\,\left[ \cos{(x)} + 1 \right] - 1 \right\} &= 0 \\ \left[ \cos{(x)} - 1 \right] \left[ 8\cos^3{(x)} + 8\cos^2{(x)} - 1 \right] &= 0 \end{align*}$

This factors further. Hint: $\displaystyle \begin{align*} \cos{(x)} = -\frac{1}{2} \end{align*}$ is another solution.
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